Find the distance between the point ${(6, 1)}$ and the line $\enspace {y = 3x + 3}\thinspace$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
First, find the equation of the perpendicular line that passes through ${(6, 1)}$ The slope of the blue line is ${3}$ , and its negative reciprocal is ${-\dfrac{1}{3}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -\dfrac{1}{3}x + b}\thinspace$ We can plug our point, ${(6, 1)}$ , into this equation to solve for ${b}$ , the y-intercept. $1 = {-\dfrac{1}{3}}(6) + {b}$ $1 = -2 + {b}$ $1 + 2 = {b} = 3$ The equation of the perpendicular line is $\enspace {y = -\dfrac{1}{3}x + 3}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(0, 3)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(6, 1)}$ and ${(0, 3)}$ gives us: $\sqrt{( {6} - {0} )^2 + ( {1} - {3} )^2}$ $= \sqrt{( 6 )^2 + ( -2 )^2} = \sqrt{40} = 2\sqrt{10}$ The distance between the point ${(6, 1)}$ and the line $\thinspace {y = 3x + 3}\enspace$ is $\thinspace2\sqrt{10}$.